EXPLANATION OF MOLYBDENUM IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) May 28 , 2015 Molybdenum is a chemical element with symbol Mo and atomic number 42. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECTING EINSTEIN). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this correct image of Molybdenum including the following ground state electron configuration: 1s22s22p63s23p63d104s2 4p64d5 5s1 . According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of molybdenum (from (E1 to E30) are the following: E1 = 7.09, E2 = 16.16, E3 = 27.13 , E4 = 46.4 , E5 = 54.49, E6 = 68.83, E7 = 125.66, E8= 143.6, E9 = 164.12, E10 = 186.4, E11 = 209.3 , E12 = 230.28, E13 = 279.1, E14 = 302.6 , E15 = 544, E16 = 570, E17 = 636, E18 = 702, E19 = 767, E20 = 833 , E21 = 902 E22 = 968, E23 = 1020 , E24 = 1082, E25 = 1263, E26 =1323 E27 = 1387, E28= 1449 E29 = 1535, E30 = 1601. Firstly we examine the - E1 = -7.09 = E(5s1) Here the E(5s1) represents the binding energy of the outermost electron (5s1) . Then we observe that -( E2 +…+ Ε6 ) = -213 = E(4d5) -( Ε7 +…+ E12 ) = -1059.36 = E(4p6) - ( E13 + E14 ) = -581.7 = E(4s2). -(E15 +…+ E24) = - 8024 = E(3d10) -(E25 +…+ E30 ) = - 8558 = E(3p6) It is of interest to note that in the absence of data (from E31 to E42 ) one can write the following theoretical relations of ionizations to the ground state energies: -( E31 + E32 ) = E(3s2) -(E33 +..+ E38) = E(2p6) -(E39 + E40) = E(2s2) -( E41 + E42) = E(1s2) Such theoretical ionization energies are analogous to the experimental values of the ionizations of copper. ( See my EXPLANATION OF COPPER IONIZATIONS ). For understanding better the ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008. ' ' EXPLANATION OF - E1 = -7.09 = E(5s1) Here the binding energy E(5s1) of the outermost electron is given by applying the Bohr formula . The charges (-41e) of the electrons (1s22s22p63s23p63d104s24p64d5 ) screen the nuclear charge (+42e) and for a perfect screening we would have an effective ζ = 1. However the electron 5s1 penetrates the 4d5 leading to the deformation of spherical electron clouds. Thus ζ > 1. Under this condition we may write E1 = 7.09 eV = -E(5s1) = - (- 13.6057 )ζ2 / n2 Since n = 5 we get ζ = 3.6 > 1 EXPLANATION OF -( Ε2 +…+ E6 ) = - 213 = E(4d5) ''' Here the E(4d5) represents the binding energy of the 5 electrons with parallel spin given by applying the Bohr formula. The charges (-36e) of the inner electrons (1s22s22p63s23p63d10 4s24p6) screen the nuclear charge (+42e) and for a perfect screening we would have ζ = 6. However the electrons of 4d5 repel the 4p6 and lead to the deformations of spherical electron clouds. Thus we must observe that ζ > 6. Under this condition we may write ( Ε2 +…+ E6 ) = 213 = -E(4d5) = - 5[(-13.6057)ζ2 / n2 Since n = 4 we get ζ = 7 > 6 ' ' '''EXPLANATION OF ( E7 +… + E12 ) = 1059.36 eV = -E(4p6) Here the E(4p6) represents the binding energy of the 6 electrons (4p6) . Note that the 6 electrons create three orbitals with electrons of opposite spin given by applying my formula of 2008. The charges (-30e) of the inner electrons (1s22s22p63s23p63d104s2) screen the nuclear charge (+42e) and for a perfect screening we would have ζ = 12. However the electrons of 4p6''' repel the 4s2 and lead to the deformation of spherical clouds. Thus we must observe that ζ > 12. Under this condition we may write ' '''E7 +..+ E12 ) = 1059.36 eV = - E(4p6) = -3+ (16.95)ζ - 4.1 / n2 Since n = 4 we may rewrite 5.1ζ2 - 3.178ζ - 1058.59 = 0 Then solving for ζ we get ζ = 14.72 > 12 , ' ' ' EXPLANATION OF ( E13 + E14 ) = 581.7 eV = -E(4s2) ''' Here the E(4s2) represents the binding energy of the two paired electrons (4s2) given by applying my formula of 2008. The charges (-28e) of the inner 28 electrons of (1s22s22p63s23p63d10) screen the nuclear charge (+42e) and for a perfect screening we would have an effective ζ = 14. However the two electrons of 4s2 penetrate the 3d10 leading to the deformation of spherical electron clouds . Thus we should have ζ > 14. Under this condition we write the following equation as ( E13 + E14 ) = 581.7 eV = - E(4s2) = - 27.21)ζ2 + (16.95) ζ - 4.1 / n2 Since n = 4 we may write 1.7 ζ2 - 1.06ζ - 581.444 = 0 Then solving for ζ we get ζ = 18.8 > 14 . Note that the two electrons of opposite spin (4s2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy. However in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008. '''EXPLANATION OF (E15 +…+E24) = 8024 eV = -E(3d10) Here the binding energy E(3d10) of the 10 electrons of 3d10 is given by applying my formula of 2008. The charges (-18e) of the inner electrons (1s22s22p63s23p6) screen the nuclear charge (+42e) and for a perfect screening we would have ζ = 24. Under this condition we may write (E15 +…+E24) = 8024 eV = -E(3d10) = -527.21)ζ2 - (16.95)ζ + (4.1) /n2 Surprisingly using n = 3 and solving for ζ we get ζ < 24, which cannot exist . In fact, the 10 electrons of the five orbitals make a complete spherical sub- shell leading to a perfect screening with ζ = 24. Thus using ζ = 24 we expect to determine n > 3. Under this condition we may write E15 +..+ E24 ) = 8024 eV = -E(3d10) = -527.21)242 - (16.95)24 + (4.1) / n2 Then solving for n we get n = 3.08 > 3 . ' ' EXPLANATION OF ( E25 + …+ E30 ) = 8558 eV = -E(3p6) Here the E(3p6) represents the binding energy of the 6 paired electrons given by applying my formula of 2008. ' '''The charges (-12e) of the twelve inner electrons like (1s22s22p63s2 ) screen the nuclear charge (+42e) and for a perfect screening we would have an effective Zeff = ζ = 30. However the 6 paired electrons ( 3p6 ) repel the 3s2 electrons and lead to the deformation of spherical electron clouds.Thus we would observe that ζ > 30. Under this condition we may write ( E25+…+ E30) = 8558 eV' = '-E(3p6)' = '''-3+(16.95)ζ - 4.1 / n2 Since n = 3 the above equation can be written as 9.07ζ2 - 5.65ζ - 8556.63 = 0 Then, solving for ζ we get ζ = 31 > 30 . Category:Fundamental physics concepts